15,000 * (1.1)^16
I don't have a calculator on me to do it but that's the working.
Answer:
it depends on the type of line but for a straight line the starting point is always the edge that is A
Answer:
See explanation
Step-by-step explanation:
Consider two triangles ABF and EDF. These triangles are two right triangles, because angles B and D are right anges.
In these triangles,
- ∠ABF ≅ ∠EDF - as two right angles;
- BF ≅ FE - given;
- ∠AFB ≅ ∠EFD - as vertical angles when two lines AD and BE itersect.
By ASA postulate (or HA postulate) these triangles are congruent, so
ΔABF ≅ ΔEDF
Congruent triangles have congruent corresponding parts, so
FA ≅ EF
Answer:
y=-5/3x+20
Step-by-step explanation:
Let the equation of the required line be represented as ![\[y=mx+c\]](https://tex.z-dn.net/?f=%5C%5By%3Dmx%2Bc%5C%5D)
This line is perpendicular to the line ![\[y=\frac{3}{5}x+10\]](https://tex.z-dn.net/?f=%5C%5By%3D%5Cfrac%7B3%7D%7B5%7Dx%2B10%5C%5D)
![\[=>m*\frac{3}{5}=-1\]](https://tex.z-dn.net/?f=%5C%5B%3D%3Em%2A%5Cfrac%7B3%7D%7B5%7D%3D-1%5C%5D)
![\[=>m=\frac{-5}{3}\]](https://tex.z-dn.net/?f=%5C%5B%3D%3Em%3D%5Cfrac%7B-5%7D%7B3%7D%5C%5D)
So the equation of the required line becomes ![\[y=\frac{-5}{3}x+c\]](https://tex.z-dn.net/?f=%5C%5By%3D%5Cfrac%7B-5%7D%7B3%7Dx%2Bc%5C%5D)
This line passes through the point (15.-5)
![\[-5=\frac{-5}{3}*15+c\]](https://tex.z-dn.net/?f=%5C%5B-5%3D%5Cfrac%7B-5%7D%7B3%7D%2A15%2Bc%5C%5D)
![\[=>c=20\]](https://tex.z-dn.net/?f=%5C%5B%3D%3Ec%3D20%5C%5D)
So the equation of the required line is ![\[y=\frac{-5}{3}x+20\]](https://tex.z-dn.net/?f=%5C%5By%3D%5Cfrac%7B-5%7D%7B3%7Dx%2B20%5C%5D)
Among the given options, option 4 is the correct one.
Mxn Matrices do not have an inverse solution for m ≠ n, so your 2x4 will not have a solution.
<span>Generally, inverting matrices is done through finding the determinant of the square matrix and then performing an operation on the matrix based on the size of the matrix. </span>
<span>Wikipedia has a good article on Matrix inversion, take a look there for more information.</span>
